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a(n) = n^7 + n^6 + n^5 + n^4 + n^3 + n^2 + n + 1.
8

%I #32 Oct 04 2024 11:18:31

%S 1,8,255,3280,21845,97656,335923,960800,2396745,5380840,11111111,

%T 21435888,39089245,67977560,113522235,183063616,286331153,435984840,

%U 648232975,943531280,1347368421,1891142968,2613136835,3559590240,4785883225,6357828776,8353082583

%N a(n) = n^7 + n^6 + n^5 + n^4 + n^3 + n^2 + n + 1.

%C a(n) = 11111111 in base n for n>0.

%H Carlos M. da Fonseca and Anthony G. Shannon, <a href="https://doi.org/10.7546/nntdm.2024.30.3.491-498">A formal operator involving Fermatian numbers</a>, Notes Num. Theor. Disc. Math. (2024) Vol. 30, No. 3, 491-498.

%H <a href="/index/Rec#order_08">Index entries for linear recurrences with constant coefficients</a>, signature (8,-28,56,-70,56,-28,8,-1).

%F a(n) = (n^8-1)/(n-1) for n != 1.

%F G.f.: 1 -x*(x^7-8*x^6-57*x^5-1016*x^4-2297*x^3-1464*x^2-191*x-8)/(x-1)^8. - _Colin Barker_, Oct 29 2012

%F E.g.f.: exp(x)*(1 + 7*x + 120*x^2 + 423*x^3 + 426*x^4 + 156*x^5 + 22*x^6 + x^7). - _Stefano Spezia_, Oct 03 2024

%e a(3) = 3280 because 11111111 base 3 = 2187+729+243+81+27+9+3+1 = 3280.

%t Table[FromDigits["11111111",n],{n,1,30}] (* or *) Table[n^7+n^6+n^5+n^4+n^3+n^2+n+1,{n,1,60}] (* _Vladimir Joseph Stephan Orlovsky_, Jan 29 2012 *)

%Y 8th row of the array A055129.

%Y Cf. A104878.

%K nonn,base,easy

%O 0,2

%A _Henry Bottomley_, Mar 23 2000

%E a(0)=1 prepended by _Alois P. Heinz_, May 04 2021